Metamath Proof Explorer


Theorem opncld

Description: The complement of an open set is closed. (Contributed by NM, 6-Oct-2006)

Ref Expression
Hypothesis iscld.1 𝑋 = 𝐽
Assertion opncld ( ( 𝐽 ∈ Top ∧ 𝑆𝐽 ) → ( 𝑋𝑆 ) ∈ ( Clsd ‘ 𝐽 ) )

Proof

Step Hyp Ref Expression
1 iscld.1 𝑋 = 𝐽
2 simpr ( ( 𝐽 ∈ Top ∧ 𝑆𝐽 ) → 𝑆𝐽 )
3 1 eltopss ( ( 𝐽 ∈ Top ∧ 𝑆𝐽 ) → 𝑆𝑋 )
4 1 isopn2 ( ( 𝐽 ∈ Top ∧ 𝑆𝑋 ) → ( 𝑆𝐽 ↔ ( 𝑋𝑆 ) ∈ ( Clsd ‘ 𝐽 ) ) )
5 3 4 syldan ( ( 𝐽 ∈ Top ∧ 𝑆𝐽 ) → ( 𝑆𝐽 ↔ ( 𝑋𝑆 ) ∈ ( Clsd ‘ 𝐽 ) ) )
6 2 5 mpbid ( ( 𝐽 ∈ Top ∧ 𝑆𝐽 ) → ( 𝑋𝑆 ) ∈ ( Clsd ‘ 𝐽 ) )